Simplification MCQ Quiz - Objective Question with Answer for Simplification - Download Free PDF

Given:

The fractions are: (4/9), (5/8), (2/3), and (3/4).

Method:

We will compare the fractions by converting them to decimals or finding a common denominator. First, let's convert each fraction to a decimal:

4/9 = 0.4444

5/8 = 0.625

2/3 = 0.6666

3/4 = 0.75

Comparing the decimals:

0.4444, 0.625, 0.6666, and 0.75

Conclusion:

The largest value is 0.75, which corresponds to the fraction 3/4.

Therefore, the largest fraction is 3/4.

Given:

\(\frac{(-2-3) ×(5+3) ÷(-2-3)}{(-6-4) ÷(-7-5)}\)

Formula Used:

Follow the order of operations (PEMDAS/BODMAS)

Calculation:

Calculate the values step by step:

(-2 - 3) = -5

(5 + 3) = 8

(-2 - 3) = -5

(-6 - 4) = -10

(-7 - 5) = -12

Now substitute back into the expression:

\(\left(-5 × 8 ÷ -5\right) ÷ \left(-10 ÷ -12\right)\)

Simplify inside the parentheses first:

-5 × 8 = -40

-40 ÷ -5 = 8

Simplify the second part:

-10 ÷ -12 = \(\frac{10}{12} = \frac{5}{6}\)

Now divide the results:

8 ÷ \(\frac{5}{6}\) = 8 × (6/5)

48/5 = 9.6

The correct answer is option 4, 9.6.

Given:

The expression is: (20 × 13) × {4 ÷ 4 × (19 - 13) / 3}

Formula Used:

Follow the BODMAS rule (Brackets, Orders, Division, Multiplication, Addition, Subtraction).

Calculation:

⇒ (20 × 13) × {4 ÷ 4 × (19 - 13) / 3}

⇒ (20 × 13) × {4 ÷ 4 × 6 / 3}

⇒ (20 × 13) × {1 × 6 / 3}

⇒ (20 × 13) × 2

⇒ 20 × 13 = 260

⇒ 260 × 2 = 520

The value of the expression is 520.

Given:

\(\frac{1}{\sqrt{9} + \sqrt{11}} + \frac{1}{\sqrt{11} + \sqrt{13}} + \cdots + \frac{1}{\sqrt{2023} + \sqrt{2025}}\) = ?

Formula used:

1a+b=b−a" id="MathJax-Element-23-Frame" role="presentation" style="position: relative;" tabindex="0">1a√+b√=b√−a−−√1a+b=b−a $\frac{1}{\sqrt{a} + \sqrt{b}} = \sqrt{b} - \sqrt{a}$

Calculation:

19+11+111+13+⋯+12023+2025" id="MathJax-Element-24-Frame" role="presentation" style="position: relative;" tabindex="0">19√+11√+111√+13√+⋯+12023√+2025√19+11+111+13+⋯+12023+2025 $\frac{1}{\sqrt{9} + \sqrt{11}} + \frac{1}{\sqrt{11} + \sqrt{13}} + \cdots + \frac{1}{\sqrt{2023} + \sqrt{2025}}$

⇒ (11−9" id="MathJax-Element-25-Frame" role="presentation" style="position: relative;" tabindex="0">11−−√−9–√11−9 $\sqrt{11} - \sqrt{9}$ ) + (13−11" id="MathJax-Element-26-Frame" role="presentation" style="position: relative;" tabindex="0">13−−√−11−−√13−11 $\sqrt{13} - \sqrt{11}$ ) + ... + (2025−2023" id="MathJax-Element-27-Frame" role="presentation" style="position: relative;" tabindex="0">2025−−−−√−2023−−−−√2025−2023 $\sqrt{2025} - \sqrt{2023}$ )

⇒ -9" id="MathJax-Element-28-Frame" role="presentation" style="position: relative;" tabindex="0">9–√9 $\sqrt{9}$ + 2025" id="MathJax-Element-29-Frame" role="presentation" style="position: relative;" tabindex="0">2025−−−−√2025 $\sqrt{2025}$

9=3" id="MathJax-Element-30-Frame" role="presentation" style="position: relative;" tabindex="0">9–√=39=3 $\sqrt{9} = 3$ , 2025=45" id="MathJax-Element-31-Frame" role="presentation" style="position: relative;" tabindex="0">2025−−−−√=452025=45 $\sqrt{2025} = 45$

⇒ 45 - 3 = 42

∴ The correct answer is option (4).

Concept Used: 

\({1 \over √n + √(n-1)} = {1 \over √ n + √ (n-1)} \times {(√n - √(n-1)) \over(√n - √(n-1))}\)

\({1 \over √n + √(n-1)} = {√n - √(n-1) \over (n - (n-1))} = √ n - √ (n-1)\)

Calculation:

\({1 \over √10 + √9} = { (√10 - √9) \over (√ 10 + √9)(√10 - √9)}= {√ 10 - √9 \over (10 - 9 )}\) = √10 - √9

Similarly, 

\({1 \over √11 + √10} = \) √11 - √10, and \({1 \over √12 + √11} = √12 - √11\) 

putting the values in the equation, 

⇒ (√10 - √9) + (√11 - √10) + (√12 -√11) + ................ + (√196 - √195) 

⇒ (-√9 + √196)

⇒  (-3 + 14) = 11

∴ The value of this expression is 11

Concebt used

a.b̅ = a.bbbbbb

a.0b̅ = a.0bbbb

Calculation

0.7 = 0.700000 ̇....

\(0.\bar7 = 0.77777 \ldots\)

\(0.0\bar7 = 0.077777 \ldots\)

\(0.\overline {07} = 0.070707 \ldots\)

Solution:

\(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}\)

= 25/2 + 37/3 + 73/6

= (75 + 74 + 73)/6

= 222/6

= 37

Shortcut Trick

\(12\frac{1}{2} + 12\frac{1}{3} + 12\frac{1}{6}\)

= 12 + 12 + 12 + (1/2 + 1/3 + 1/6)

= 36 + 1 = 37

Solution:

We have to follow the BODMAS rule

Calculation:

\(? = \sqrt[5]{{{{\left( {243} \right)}^2}}}\)

\(⇒ ? = (243)^{\frac{2}{5}}\)

\(⇒ ? = (3 × 3 × 3 × 3 × 3)^{2⁄5}\)

\(⇒ ? = (3^5)^{2⁄5}\)

⇒ ? = 3²

Formula used:

(a + b)² = a² + b² + 2ab

Calculation:

Given expression is:

\(\sqrt {8\; + \;2\sqrt {15} \;} \)

⇒  \(\sqrt {5\; + \;3\; + \;2\times \sqrt 5 \times \sqrt 3 \;} \)

⇒  \(\sqrt {{{(\sqrt 5 )}^2}\; + \;{{\left( {\sqrt 3 } \right)}^2}\; + \;2 \times \sqrt 5 \times \sqrt 3 \;} \)

⇒  \(\sqrt {{{\left( {\;\sqrt 5 \; + \;\sqrt 3 \;} \right)}^2}\;} \)

⇒  \(\sqrt 5 + \sqrt 3 \)

\(\sqrt {{{\left( {0.65} \right)}^2} - {{\left( {0.16} \right)}^2}} \)

Since,

a2 - b2 = (a - b) ( a + b)

\(\begin{array}{l} \Rightarrow \sqrt {\left( {0.65 + 0.16} \right)\left( {0.65 - 0.16} \right)} \\ \Rightarrow \sqrt {\left( {0.81} \right)\left( {0.49} \right)} \\ \Rightarrow \sqrt {\left( {0.9} \right)\left( {0.9} \right) \times \left( {0.7} \right)\left( {0.7} \right)} \end{array}\)

⇒ 0.9 × 0.7 = 0.63

Concept:

We can find √x using the factorisation method.

Calculation:

√[(10 + √25) (12 - √49)]

⇒ √[(10 + 5)(12 – 7)]

⇒ √(15 × 5)

⇒ √(3 × 5 × 5)

⇒ 5√3

Given,

2³ × 3⁴ × 1080 ÷ 15 = 6^x

⇒ 2³ × 3⁴ × 72 = 6^x

⇒ 2³ × 3⁴ × (2 × 6²) = 6^x

⇒ 2⁴ × 3⁴ × 6² = 6^x

⇒ (2 × 3)⁴ × 6² = 6^x           [∵ x^m × y^m = (xy)^m]

⇒ 6⁴ × 6² = 6^x

⇒ 6^{(4 + 2)} = 6^x

⇒ x = 6

Given:

√3n = 729

Formulas used:

(x^a)^b = x^ab

If x^a = x^b then a = b 

Calculation:

√3n = 729

⇒ √3n = (3²)³

⇒ (3n)^1/2 = (32)3

⇒ (3n)1/2 = 3⁶

⇒ n/2 = 6 

∴  n = 12 

Given expression,

(81.84 + 118.16) ÷ 5³ = 1.2 × 2 + ?

⇒ 200 ÷ 5³ = 1.2 × 2 + ?

⇒ 200 ÷ 125 = 1.2 × 2 +?

⇒ 1.6 = 2.4 + ?

⇒ ? = -0.8